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Continuous random variables
So far we have been concentrating on discrete random variables, whose distributions are not continuous. Now we deal with the so-called continuous random variables. A random variable X is called a continuous random variable if its distribution function F is a continuous function on R, or equivalently, if
P(X = x) = 0, for every x ∈ R.
Of course there are random variables which are neither discrete nor continuous. But in this course we will concentrate on discrete random variables and continuous random variables. Among continuous random variables, we will concentrate on a subclass.
Definition 1 A random variable X is called an absolutely continuous random variable if there is a nonnegative function f on R such that
P(X ≤ x) =
∫ (^) x
−∞
f (t)dt, for every x ∈ R.
The function f in the definition above is called the probability density of the absolutely con- tinuous random variable X and it must satisfy ∫ (^) ∞
−∞
f (t)dt = 1.
Definition 2 A nonnegative function f on R satisfying ∫ (^) ∞
−∞
f (t)dt = 1.
is called a probability density function.
It can be shown that if f is a probability density function, then then there exists an absolutely continuous random variable X with f as its probability density function. Given a nonnegative function g on R such that
−∞ g(t)dt^ =^ c^ is a finite positive number, then the function f (x) = (^1) c g(x) is a probability density function. From Definition 1, we can easily see that an absolutely continuous random variable is a contin- uous random variable. However, there are continuous random variables which are not absolutely continuous and we call these random variables singularly continuous random variables. In this course, we will not deal with singularly continuous random variables. If X is an absolutely continuous random variable with density f , then its distribution function is given by
F (x) =
∫ (^) x
−∞
f (t)dt, for every x ∈ R.
Conversely, if X is an absolutely continuous random variable with distribution function F , then its density function is given by f (x) = F ′(x)
at those points x where F is differentiable and f (x) = 0 at those points x where F is not differen- tiable. If X is an absolutely continuous random variable with density f , then for any real numbers a < b we have
P(X ∈ (a, b)) = P(X ∈ [a, b]) =
∫ (^) b
a
f (t)dt.
Example 3 A point is chosen randomly from the unit disk so that each point in the disk is equally likely. Let X the distance of the point from the center of the disk. Then the distribution function of X is given by
F (x) =
0 , x ≤ 0 x^2 , 0 < x ≤ 1 1 , x > 1.
This random variable is an absolutely continuous random variable with density
f (x) =
2 x, x ∈ (0, 1) 0 , otherwise.
Example 4 Suppose that X is an absolutely continuous random variable whose density function is given by
f (x) =
C(4x − 2 x^2 ), 0 < x < 2 0 , otherwise.
(a) What is the value of C? (b) Find P(X > 1). Solution. (a) Since f is a probability distribution we must have ∫ (^) ∞
−∞
f (x)dx = C
0
(4x − 2 x^2 )dx = 1.
Since
0 (4x^ −^2 x
(^2) )dx = 8 3 , we have^ C^ =^
8
(b) P(X > 1) =
1 f^ (x)dx^ =^
3 8
1 (4x^ −^2 x
(^2) )dx = 1
Definition 5 If X is an absolutely continuous random variable with density f , then the expectation of X is defined as
E[X] =
−∞
xf (x)dx.
Example 6 Suppose that X is an absolutely continuous random variable with density given by
f (x) =
2 x, x ∈ (0, 1) 0 , otherwise.
Then the expectation of X is
E[X] =
−∞
xf (x)dx =
0
2 x^2 dx =^2 3
Example 12 Buses arrive at a specified bus stop at 15 minutes intervals starting at 7 am. That is, they arrive at 7, 7:15, 7:30, and so on. If a passenger arrives at a stop at a time that is uniformly distributed between 7 and 7:30, find the probability that he waits (a) less than 5 minutes for a bus; (b) more than 10 minutes for a bus.
Example 13 A stick of length 1 is split at a point X that is uniformly distributed over (0, 1). Suppose that p is a point in (0, 1). Determine the expected length of the piece that contains the point p.
Now we introduce the normal random variables. Consider the function
g(x) = e−x^2 /^2 , x ∈ R.
It is easy to see that the integral
c =
−∞
g(x)dx
converges and thus c is a finite positive number. There is no simple formula for the antiderivative of g and so we can not evaluate the value of c by the fundamental theorem of calculus. The easiest way to evaluate c is by a very special trick in which we we write c as a two-dimensional integral and introduce polar coordinates.
c^2 =
−∞
e−x^2 /^2 dx
−∞
e−y^2 /^2 dy =
−∞
−∞
e−(x^2 +y^2 )/^2 dxdy
0
(∫ (^) π
−π
e−r
(^2) / 2 rdθ
dr = 2π
0
re−r
(^2) / 2 dr
= − 2 πe−r^2 /^2 |∞ 0 = 2π.
Therefore the function f (x) = √^1 2 π
e−x^2 /^2 , x ∈ R
is a probability density and it is called the standard normal density. An absolutely continuous random variable whose density is a standard normal density is called a standard normal random variable. There is no simple formula for the distribution function of a standard normal random variable. It is traditional to denote the distribution function of a standard normal random variable by
Φ(x) = √^1 2 π
∫ (^) x
−∞
e−t^2 /^2 dt, x ∈ R.
One can find the values of Φ(x) by tables given in probability textbooks or by computer. It is easy to check that Φ satsfies the following relation
Φ(−x) = 1 − Φ(x), x ∈ R.
Suppose that X is a standard normal random variable, μ is a real number and σ > 0. Let consider the random variable Y = μ + σX. For any y ∈ R,
P(Y ≤ y) = P(μ + σX ≤ y) = P(X ≤ (y − μ)/σ)
√^1
2 π
∫ (^) (y−μ)/σ
−∞
e−t
(^2) / 2 dt.
By the chain rule and the fundamental of calculus we know that Y is an absolutely continuous random variable whose density is given by
f (y) = 1 σ
2 π
e(y−μ)^2 /^2 σ^2 , y ∈ R.
The probability density above is called a normal density with parameters μ and σ^2 , and will be denoted by n(μ, σ^2 ) or n(y; μ, σ^2 ). An absolutely continuous random variable whose density is a normal density n(μ, σ^2 ) is called a normal random variable with parameters μ and σ^2. The standard normal density is simply a normal density with parameters 0 and 1 n(0, 1). Thus we have shown that if X is a standard normal random variable, μ is a real number and σ > 0, then the random variable Y = μ + σX is a normal random variable with parameters μ and σ^2. Similar to the paragraph above, we can show that, if X is a normal random variable with parameters μ and σ^2 , then Z = (X − μ)/σ is a standard normal random variable. This fact will be very useful in finding the probabilities involving general normal random variables.
Theorem 14 If X is a normal random variable with parameters μ and σ^2 , then
E[X] = μ, Var(X) = σ^2.
Proof Since X can be written as μ + σZ, where Z is a standard normal random variable, so we only need to prove the theorem when X is a standard normal random variable. So we assume that X is a standard normal random variable. Since the function x √^12 π e−x^2 /^2 is an odd function, we get that
E[X] =
−∞
x √^1 2 π
e−x^2 /^2 dx = 0.
Using integration by parts (with u = x, dv = xe−x^2 /^2 ) we get
Var(X) = E[X^2 ] = √^1 2 π
−∞
x^2 e−x^2 /^2 dx
2 π
−xe−x^2 /^2 |∞−∞ +
−∞
e−x^2 /^2 dx
√^1
2 π
−∞
e−x
(^2) / 2 dx = 1.
Example 15 If X is a normal random variable with parameters μ = 3 and σ^2 = 9, find (a) P(2 < X < 5); (b) P(X > 0); (c) P(|X − 3 | > 6). Solution (a).
P(2 < X < 5) = P
X − 3
3 <^
= P
X − 3
3 <^
3 )^ −^ Φ(−^
3 )^ ≈^.^3799
So the distribution function of X is given by
F (x) =
1 − e−λx^ x ≥ 0 0 x < 0
Theorem 18 Suppose that X is an exponential random variable with parameter λ, then E[X] = (^1) λ and Var(X) = (^) λ^12.
Proof For any positive integer n, using integration by parts, we have
E[Xn] =
0
xnλe−λxdx
= −xne−λx|∞−∞ +
0
nxn−^1 e−λxdx
= n λ
0
xn−^1 λe−λxdx = n λ
E[Xn−^1 ].
Letting n = 1 and then n = 2, we get E[X] = (^) λ^1 and E[X^2 ] = (^2) λ E[X] = (^) λ^22. Thus Var(X) = (^) λ^12.
Example 19 Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 101. If someone arrives immediately ahead of you at a public telephone booth, find the probability that you have to wait (a) more than 10 minutes; (b( between 10 and 20 minutes.
We say that a positive random variable X is memoryless if
P(X > s + t|X > t) = P(X > s), for all s, t > 0.
It is very easy to check that any exponential random variable is memoryless. In fact, we can prove that if a positive random variable X is memoryless in the sense above, it must be an exponential random variable. This suggests that exponential random variables are very useful in modeling lifetimes of certain equipment when the memoryless property is approximately satisfied. To introduce the next type of random variable, we recall the definition of Gamma function. For any α > 0, the Gamma function is defined by
Γ(α) =
0
xα−^1 e−xdx.
There is no explicit formula for this function. One can easily see that Γ(1) = 1. In fact, we can easily find a formula for Γ(n) for any positive integer n. Before this, we observe the following
Proposition 20 For any α > 0 , we have Γ(α + 1) = αΓ(α).
proof
Γ(α + 1) =
0
xαe−xdx
= −xαe−x|∞ 0 +
0
αxα−^1 e−xdx = αΓ(α).
Using this proposition, we immediately get Γ(n) = (n − 1)!. One can easily check that, for any α > 0 and λ > 0, the function
f (x) =
{ (^) λα Γ(α) x
α− (^1) e−λx (^) x > 0 0 x ≤ 0
is a probability density function and it is called a Gamma density with parameters α and λ. For any α > 0 and λ > 0, an absolutely continuous random variable is called a Gamma random variable with parameters α and λ if its density is a Gamma density with parameters α and λ. A Gamma random variable with parameters α = 1 and λ is simply an exponential random variable with parameter λ.
Theorem 21 Suppose that X is a Gamma random variable with parameters α and λ, then E[X] = α λ and^ Var(X) =^
α λ^2. Proof
E[X] = (^) Γ(^1 α)
0
λxe−λx(λx)α−^1 dx
= 1 λΓ(α)
0
λe−λx(λx)αdx
= Γ( λαΓ(^ + 1)α) = αλ.
Similarly, we can find E[X^2 ] and then find that Var(X) = (^) λα 2.
Example 22 Suppose that X is a normal random variable with parameters μ = 0 and σ^2. Find the density of Y = X^2.
Solution Let F be the distribution function of X and f the density function of X. Y is a nonnegative random variable. For any y > 0, we have
P(Y ≤ y) = P(X^2 ≤ y) = P(−√y ≤ X ≤ √y) = F (√y) − F (−√y).
Thus the density of Y is
g(y) =
1 2 √y (f^ (
√y) + f (−√y)) y > 0
0 y ≤ 0.
Plugging in the expression for f , we get that the density of Y is
g(y) =
1 σ√ 2 πy e
−y/(2σ^2 ) (^) y > 0
0 y ≤ 0.
This example shows that, if X is a normal random variable with parameters μ = 0 and σ^2 , then X^2 is a Gamma random variable with parameters α = 12 and λ = (^21) σ 2. By comparing (0.1) with definition of Gamma density we see that
Γ(
2 ) =^
π.
